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05/31/2004 03:20:25 PM · #1 |
With the Deep DOF challenge I have been looking over the calculations for the hyperfocal distance and for calculating DOF in general. For the math I am using the second page of this Vivid Light article.
(warning, math ahead)
I am attempting to calculate the circle of confusion for my camera in order to create the most accurate chart possible. The article states the CoC is calculated by 1/(1300/D), where D is the diagonal of the chip in mm or D=sqrt(a*a+b*b) where a and b are the lengths of the sides of the format.
This is where I become confused. In the manual for my camera, Olympus C720UZ, it says the image pickup device is a 1/2.5" CCD. Assuming this is the correct value for the diagonal, I proceed to convert from inches to mm; 1/2.5" = 0.4" = 10.16mm. Using this value in the equation I get CoC = 1/(1300/10.16) = 0.007815; which is quite a bit lower than the suggested 0.02 earlier in the article.
I know, math does not translate well in plain text, and becomes hard to read, but could those of you inclined to do so look it over and correct me if my calculations are inaccurate ... or if I used the wrong starting value.
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The second issue I am having is with the focal length used in the calculations of the DOF; again from the manual, my lens is "6.4mm to 51.2mm, f2.8 to f3.4, 10 elements in 7 groups (equivalent to 44mm to 320 mm lens on 35mm camera)."
Ignoring for the moment that I have no idea what they are referring to with the "10 elements in 7 groups" part, do I use the 6.4mm to 51.2mm as my focal lengths or the 44mm to 320mm? or is that equivalence accounted for in the difference in the value for the CoC? Do I even understand the term focal length well enough to be looking at the right numbers in the manual?
Thanks for looking, and I hope my confusion is not contagious. ;)
David |
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05/31/2004 03:54:33 PM · #2 |
There's alot of, er, confusion, if you will, regarding CoC. And there is room for debate as to how it "should" be determined. Here's my interpretation:
Calculate the CoC using the Nyquist limit for your camera. That's simply twice the pixel pitch. You can calculate this easily: you know the sensor height and width (because you know the diagonal and the aspect ratio), and you also know how many pixel columns and rows are present.
The above does assume that the lens system is capable of providing enough resolution to approach the limits of the sensor performance. This is approximately true for most good cameras, at least stopped down a bit.
edit:
O, forgot, yes, you use the actual focal length numbers, not the "35mm equivalents". That's very important.
BTW, "10 elements in 7 groups" means that there are 10 separate pieces of glass in your cam's lens system, and they are in 7 discrete groups (some may be cemented together in pairs, refered to as "doublets")
Message edited by author 2004-05-31 15:59:20.
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05/31/2004 04:10:18 PM · #3 |
OK, so if a train is leaving Pittsburg at 1:00 pm, and another train is leaving San Francisco at 1:30 pm, how long before this Circle of Confusion reaches the DFW Metroplex area? :o) LOL
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05/31/2004 04:30:29 PM · #4 |
Also, see this thread for additional information.
Often, a rough approximate CoC value is used for simplicity; the 0.02 value usually assumes a fairly large sensor (as found on many dSLRs). However, it really depends on your final print size (i.e. enlargement factor), viewing distance, and desired print quality. Most CoC values assume a final print size of 8x10 viewed from 10 inches with a quality of somewhere between 3 and 4 lp/mm (i.e. between 150 and 200 ppi). Don't forget that you have to take your crop factor into account. |
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05/31/2004 04:37:43 PM · #5 |
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05/31/2004 04:38:22 PM · #6 |
Originally posted by laurielblack:
OK, so if a train is leaving Pittsburg at 1:00 pm, and another train is leaving San Francisco at 1:30 pm, how long before this Circle of Confusion reaches the DFW Metroplex area? :o) LOL |
it all depends on the trains running away or approaching each other ;-)
David, the number stated as size of the CCD isn't, oddly enough, its actual diagonal size. You can find some advise at DPReview
Be ready for some metric math ;-)
Carlos |
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05/31/2004 04:45:22 PM · #7 |
This is much easier. DOFMaster
).005 is the calculation it comes up with for your camera. I got the download on my PDA and there is another called pCam that is more fun to use.
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05/31/2004 04:45:44 PM · #8 |
Originally posted by Britannica:
This is where I become confused. In the manual for my camera, Olympus C720UZ, it says the image pickup device is a 1/2.5" CCD. Assuming this is the correct value for the diagonal, I proceed to convert from inches to mm; 1/2.5" = 0.4" = 10.16mm. Using this value in the equation I get CoC = 1/(1300/10.16) = 0.007815; which is quite a bit lower than the suggested 0.02 earlier in the article.
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I remember reading a value of 0.02 for the Canon 300D, which has a much bigger sensor than most compacts. My Sony F717 is supposed to have a CoC of 0.008, which is awfully close to your calculated value, so you probably have done your math correctly. :-)) |
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05/31/2004 05:14:11 PM · #9 |
Here's a fairly brief, but useful, set of formulas:
depth of field equations
Note that he has a few other useful pages, including chart makers and a CoC calculator. Unfortunately, he hasn't updated it for digital formats. (I would use 0.004 instead of 0.005 for your CoC though).
For something more comprehensive, see this page
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06/01/2004 02:18:17 AM · #10 |
Thanks to all who took the time to clue me in and point me to new references.
I am beginning to think that this is one of those things in life that are made more complex than they absolutely need to be. ;)
Now, if you will excuse me, I am going to go happily bury myself in articles and equations for the rest of the night.
David
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